Angular momentum of a 2p orbital

In spherical coordinates, $x+y+2z=r\,(\sin\theta\cos\varphi+\sin\theta\sin\varphi+2\cos\theta)$. Let us calculate the angular integral of its square. By symmetry, $\int(x/r)^2d\Omega=\int(y/r)^2d\Omega=\int(z/r)^2d\Omega=4\pi/3$, and cross terms such as $\int xy\,d\Omega$ vanish because they change sign under a symmetry that reverses the sign of only one coordinate. Thus

$$\int(x+y+2z)^2\,d\Omega=r^2(1+1+4)\frac{4\pi}{3}=8\pi r^2 .$$

The normalisation condition is therefore

$$1=N^2\int_0^\infty8\pi r^2\,e^{-r/a}\,r^2dr=8\pi N^2\times4!\,a^5=192\pi a^5N^2, \qquad N=\frac{1}{\sqrt{192\pi\,a^5}}=\frac{1}{8\sqrt{3\pi}\,a^{5/2}}.$$

We have $x\pm iy=r\sin\theta\,e^{\pm i\varphi}=\mp r\sqrt{8\pi/3}\,Y_{1,\pm1}$ and $z=r\cos\theta=r\sqrt{4\pi/3}\,Y_{10}$. Let $c=r\sqrt{4\pi/3}$, so that $r\sqrt{8\pi/3}=\sqrt2\,c$. Taking the half-sum and half-difference gives

$$x=\frac{(x+iy)+(x-iy)}{2}=\frac{c}{\sqrt2}\bigl(Y_{1,-1}-Y_{11}\bigr), y=\frac{(x+iy)-(x-iy)}{2i}=\frac{ic}{\sqrt2}\bigl(Y_{11}+Y_{1,-1}\bigr), \qquad z=c\,Y_{10}.$$

It follows that

$$x+y+2z=c\left[\frac{1+i}{\sqrt2}\,Y_{1,-1}+2\,Y_{10}+\frac{-1+i}{\sqrt2}\,Y_{11}\right].$$

Since the radial factor $r\,e^{-r/(2a)}$ is proportional to $R_{21}$, the function $\psi$ is a combination of the three states $\psi_{21m}$, with coefficients proportional to

$$\frac{1+i}{\sqrt2}\ (m=-1),\qquad 2\ (m=0),\qquad \frac{-1+i}{\sqrt2}\ (m=1).$$

The three states $\psi_{21m}$ have the same energy $E_2=-13{,}6/4\simeq-3{,}4$ eV and the same value $\hbar^2\ell(\ell+1)=2\hbar^2$ of $\hat L^2$. Any combination of these states is therefore still an eigenstate of $\hat H$ and $\hat L^2$: an energy measurement gives $E_2$ with certainty, and a measurement of $\hat L^2$ gives $2\hbar^2$ with certainty.

The possible outcomes are $-\hbar$, $0$ and $\hbar$. Since the states $\psi_{21m}$ are orthonormal, the probabilities are proportional to the squared moduli of the coefficients found above:

$$\Bigl|\frac{1+i}{\sqrt2}\Bigr|^2=1,\qquad |2|^2=4,\qquad\Bigl|\frac{-1+i}{\sqrt2}\Bigr|^2=1,$$

whose sum is $6$. Thus

$$P(L_z=-\hbar)=\frac16,\qquad P(L_z=0)=\frac23,\qquad P(L_z=\hbar)=\frac16 .$$

Note that the sum of the squared moduli is $6$, as is the factor $1+1+4$ that appeared in the normalisation: this is a useful check. The expectation value is $\langle\hat L_z\rangle=-\hbar\cdot\frac16+0\cdot\frac23+\hbar\cdot\frac16=0$. This result could have been anticipated: the function $\psi$ is real, and the operator $\hat L_z=-i\hbar\,\partial_\varphi$ has a purely imaginary expectation value for a real function, which must vanish because it must also be real.

By the projection postulate, the state is projected onto the eigenspace of $\hat L_z$ associated with $\hbar$, intersected here with the level $n=2$, $\ell=1$, which is spanned by $\psi_{211}$. After normalisation, the state is $\psi_{211}$, up to a global phase:

$$\psi_{211}(\mathbf r)=R_{21}(r)\,Y_{11}(\theta,\varphi)\propto-\,r\sin\theta\,e^{i\varphi}e^{-r/(2a)}=-(x+iy)\,e^{-r/(2a)}.$$

This is not a real function: its phase rotates with the angle $\varphi$ about the $z$ axis, corresponding to a non-zero, well-defined component of angular momentum along $z$.